SME
Heat Transfer · Problem type
Centre of a plane wall (one-term solution)
Use this for “An 80 mm thick slab (L = 40 mm, α = 1.2 × 10⁻⁵ m²/s) at 200 °C is dropped in 20 °C fluid with Bi = 1 (C₁ = 1.1191, ζ₁ = 0.8603). Find the centre temperature after 300 s.”
Chart
T₀(2.25) ≈ 58.1
Level T∞ = 20 doesn't meet the curve in the window shown
This is an example. Type your own values to start a new problem.
Fo
=
α (m²/s)
×
t (s)
(
L (m)
)
2
T₀ (°C)
=
T∞ (°C)
+
(
Tᵢ (°C)
−
T∞)
×
C₁
×
e
−
2
×
Fo
Bi
Biot number, hL ÷ k (from ζ₁)
worked out (not typed)
Formulas
Bi
=
ζ₁
×
tan(
ζ₁
)
0.9999 = 0.8603 × tan(0.8603)
Fo
=
α
×
t
L
2
2.25
=
1.2 ×
10
-5
×
300
0.04
2
T₀
=
T∞
+
(Tᵢ
−
T∞)
×
C₁
×
e
−
ζ₁
2
× Fo
58.1002
=
20
+
(200
−
20)
×
1.1191
×
e
−
0.8603
2
× 2.25
Assumptions
•
The wall is heated or cooled on both faces, so L is half its thickness.
•
h, k and α stay the same; the fluid stays at T∞.
•
C₁ and ζ₁ are read from the one-term table at the wall’s Bi = hL ÷ k.
Step-by-step
WE KNOW
t = 300 s, L = 0.04 m, α = 1.2 × 10⁻⁵ m²/s, C₁ = 1.1191, ζ₁ = 0.8603, Tᵢ = 200 °C, T∞ = 20 °C
FIND
Biot number, hL ÷ k (from ζ₁) (Bi), Fourier number (Fo), centre temperature (T₀)
Step 1 · Find Biot number, hL ÷ k (from ζ₁): Bi
Use Bi = ζ₁ × tan(ζ₁)
ζ₁
is
the
first
root
of
ζ
tan
ζ
=
Bi,
so
the
table
row
it
came
from
has
this
Biot
number.
Bi = 0.8603 × tan(0.8603)
Bi = 0.8603 × 1.1623
Bi = 0.9999
Step 2 · Find Fourier number (Fo)
Use Fo = α × t ÷ L²
The
Fourier
number
measures
the
time
against
how
long
heat
takes
to
cross
L
.
Fo
=
1.2 ×
10
-5
×
300
0.04
2
Fo
=
(1.2
×
10
-5
)
×
300
0.0016
Fo
=
0.0036
0.0016
Fo = 2.25
Step 3 · Find centre temperature (T₀)
Use T₀ = T∞ + (Tᵢ − T∞) × C₁ × e−ζ₁² × Fo
The
centre
keeps
the
share
C₁
e
−
ζ₁
2
Fo
of
the
starting
gap
to
the
fluid;
add
it
to
T∞.
T₀
=
20
+
(200
−
20)
×
1.1191
×
e
−
0.8603
2
× 2.25
T₀
=
20
+
180
×
1.1191
×
e
−0.74012 × 2.25
T₀
=
20
+
180
×
1.1191
×
e
−1.6653
T₀
=
20
+
180
×
1.1191
×
0.18913
T₀
=
20
+
201.438
×
0.18913
T₀
=
20
+
38.098
T₀
=
58.1002
°C
Check
Put the numbers back in. Both sides should match.
0.9999 = 0.8603 × tan(0.8603) ✓
2.25
=
1.2 ×
10
-5
×
300
0.04
2
✓
58.1002
=
20
+
(200
−
20)
×
1.1191
×
e
−
0.8603
2
× 2.25
✓
Topic check
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